Let $g(x)=x^6-3x^5$. For what value of $x$ does $g$ have a relative minimum ? Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{2}{5}$ (Choice B) B $3$ (Choice C) C $0$ (Choice D) D $\dfrac{5}{2}$
Solution: We can find the relative extrema (i.e. minima and maxima) of $g$ by looking for the intervals where its derivative $g'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. The derivative of $g$ is $g'(x)=3x^4(2x-5)$. $g'(x)=0$ for $x=0,\dfrac{5}{2}$. Since $g'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=0$ and $x=\dfrac{5}{2}$. $g$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into three intervals: $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $(-\infty,0)$ $(0,\frac{5}{2})$ $\frac{5}{2}$ $(\frac{5}{2},\infty)$ Let's evaluate $g'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $g'(x)$ Verdict $(-\infty,0)$ $x=-1$ $g'(-1)=-21<0$ $g$ is decreasing $\searrow$ $(0,\dfrac{5}{2})$ $x=1$ $g'\left(1\right)=-9<0$ $g$ is decreasing $\searrow$ $(\dfrac{5}{2},\infty)$ $x=3$ $g'(3)=243>0$ $g$ is increasing $\nearrow$ Now let's look at the critical points: $x$ Before After Verdict $0$ $\searrow$ $\searrow$ Not an extremum $\dfrac{5}{2}$ $\searrow$ $\nearrow$ Minimum Now we can see that $g$ has a relative minimum at $x=\dfrac{5}{2}$.